The linear mass density of a rod of length L | vedclass

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(D) The position of the centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.Given the linear mass density $\lambda =

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$\frac{3L}{4}$

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(D) The position of the centre of mass $x_{cm}$ is given by the formula $x_{cm} = \frac{\int x dm}{\int dm}$.Given the linear mass density $\lambda = kx^2$,the mass element is $dm = \lambda dx = kx^2 dx$.Substituting these into the formula:$x_{cm} = \frac{\int_{0}^{L} x (kx^2 dx)}{\int_{0}^{L} kx^2 dx} = \frac{k \int_{0}^{L} x^3 dx}{k \int_{0}^{L} x^2 dx}$.Evaluating the integrals:$x_{cm} = \frac{[x^4/4]_{0}^{L}}{[x^3/3]_{0}^{L}} = \frac{L^4/4}{L^3/3} = \frac{L^4}{4} 	imes \frac{3}{L^3} = \frac{3L}{4}$.

The linear mass density of a rod of length $L$ varies as $\lambda = kx^2$,where $k$ is a constant and $x$ is the distance from one end. The position of the centre of mass of the rod is:

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